transportation problemkukg.bme.hu/.../2019/04/07_transportation-problem.pdf2019/04/07 · northwest...

BME FACULTY OF TRANSPORTATION ENGINEERING AND VEHICLE ENGINEERING32708-2/2017/INTFIN COURSE MATERIAL SUPPORTED BY EMMI

Dr. Tibor SIPOS Ph.D.Dr. Árpád TÖRÖK Ph.D.Zsombor SZABÓ2019

TRANSPORTATION PROBLEM

2

Content

Excersises

Streamlined Simplex Method

Introduction

3

• Original applications of the linear programming (LP) method

• Basic problem– a company manufactures products at 𝑚 source location (𝑎𝑖, 𝑖 = 1,… ,𝑚)

– demand for the product is distributed among 𝑛 differentabsorption locations (𝑏𝑗, 𝑗 = 1,… , 𝑛)

• Objective: deliver products from source locations toabsorption locations at minimum cost– 𝑐𝑖𝑗: delivery cost of one unit from source location 𝑖 to

absorption location 𝑗

– 𝑐𝑖𝑗𝑥𝑖𝑗: delivery cost of 𝑥𝑖𝑗 units from source location 𝑖 toabsorption location 𝑗

Basic of the transportation method

4

min 𝑍 =

𝑖=1

𝑚

𝑗=1

𝑛

𝑐𝑖𝑗𝑥𝑖𝑗

𝑗=1

𝑛

𝑥𝑖𝑗 = 𝑎𝑖 ∀𝑖

𝑖=1

𝑚

𝑥𝑖𝑗 = 𝑏𝑗 ∀𝑗

𝑖=1

𝑚

𝑎𝑖 =

𝑗=1

𝑛

𝑏𝑗 𝑎𝑛𝑑 𝑎𝑖 ≥ 0, 𝑏𝑗 ≥ 0

Linear Programming Method

5

• Theorem: For transportation problems where every 𝑎𝑖 and 𝑏𝑗 have an integer value, all the

basic variables (allocations) in every basic feasible (BF) solution (including an optimal one) also have integer values

Integer Solutions Property

6

Content

Excersises


Introduction

7

Content – Streamlined Simplex Method

Distribution Method

Finding the Initial Solution

Introduction

8


• Set up of the tabularform

• Real life examplesσ𝑖=1𝑚 𝑎𝑖 = σ𝑗=1

𝑛 𝑏𝑗– not always fulfilled– dummy supply or

destination has needed to introduce

• There are two steps– Finding the initial

solution– Distribution method

j

i

c11 c12 … c1j … c1n a1

c21 c22 … c2j … c2n a2

… … … … … … …

ci1 ci2 … cij … cin ai

… … … … … … …

cm1 cm2 … cmj … cmn am

b1 b2 … bj … bn

…

i

…

m

1

2

1 2 … j … n

9

Example problem

6 3 5 2 7 200

3 7 4 4 1 80

5 2 3 1 6 130

3 5 2 3 2 90

30 210 60 80 120

cij

3

4

1

2

1 2 3 4 5

10


Distribution Method


Introduction

11

• The goal is not to calculate the best solution, butto calculate an initial solution

• Methods

– Northwest corner route

– Minimum Costs Method (Dantzig Method)

– Vogel Initial Programming Method (The Biggest Difference Method)

Finding the initial solution

12

Northwest Corner Method

• The maximum possible flow is needto be programmedon the top left corner(cell 11)

– If 𝑏1 > 𝑎1 cell 21 is need to be chosen

– If 𝑏1 < 𝑎1 cell 12 is need to be chosen

30 170 0 0 0 200

0 40 40 0 0 80

0 0 20 80 30 130

0 0 0 0 90 90

30 210 60 80 120

3

4

1

2

xij 1 2 3 4 5

30 ∗ 6 + 170 ∗ 3 + 40 ∗ 7 + 40 ∗ 4 + 20 ∗ 3 + 80 ∗ 1 + 30 ∗ 6 + 90 ∗ 2 = 1630

13

0 0 0 0 0 200

0 0 0 0 0 80

0 0 0 0 0 130

0 0 0 0 0 90

30 210 60 80 120

1

2

3

4

1 2 3 4 5xij


14


30 0 0 0 0 170

0 0 0 0 0 80

0 0 0 0 0 130

0 0 0 0 0 90

0 210 60 80 120

3 4 5

1

xij 1 2

2

3

4

15


30 170 0 0 0 0

0 0 0 0 0 80

0 0 0 0 0 130

0 0 0 0 0 90

0 40 60 80 120

3

4

4 5

1

2

xij 1 2 3

16


30 170 0 0 0 0

0 40 0 0 0 40

0 0 0 0 0 130

0 0 0 0 0 90

0 0 60 80 120

1

2

3

4

1 2 3 4 5xij

17


30 170 0 0 0 0

0 40 40 0 0 0

0 0 0 0 0 130

0 0 0 0 0 90

0 0 20 80 120

3

4

4 5

1

2

xij 1 2 3

18


30 170 0 0 0 0

0 40 40 0 0 0

0 0 20 0 0 110

0 0 0 0 0 90

0 0 0 80 120

1

2

3

4

1 2 3 4 5xij

19


30 170 0 0 0 0

0 40 40 0 0 0

0 0 20 80 0 30

0 0 0 0 0 90

0 0 0 0 120

3

4

4 5

1

2

xij 1 2 3

20


30 170 0 0 0 0

0 40 40 0 0 0

0 0 20 80 30 0

0 0 0 0 0 90

0 0 0 0 90

1

2

3

4

1 2 3 4 5xij

21


30 170 0 0 0 0

0 40 40 0 0 0

0 0 20 80 30 0

0 0 0 0 90 0

0 0 0 0 0

3

4

4 5

1

2

xij 1 2 3

22

Minimum Costs Method (Dantzig Method)

• Task: program themaximum possibleflows on source-absorption locationrelations, wherecosts are minimal

• The order:– Red– Orange– Green– Purple

30 160 0 0 1040

0

0 0 0 0 80 0

0 50 0 80 050

0

0 0 60 0 30 0

0160

00 0

40

10

3

4

1

2

xij 1 2 3 4 5

30 ∗ 6 + 160 ∗ 3 + 10 ∗ 7 + 80 ∗ 1 + 50 ∗ 2 + 80 ∗ 1 + 60 ∗ 2 + 30 ∗ 2 = 1170

23

06 03 05 02 07 200

03 07 04 04 01 80

05 02 03 01 06 130

03 05 02 03 02 90

30 210 60 80 120

1

2

3

4

1 2 3 4 5xij


24

06 03 05 02 07 200

03 07 04 04 801 0

05 02 03 801 06 50

03 05 02 03 02 90

30 210 60 0 40

1

xij 1 2

2

3

4

3 4 5


25


06 03 05 02 07 200

03 07 04 04 801 0

05 502 03 801 06 0

03 05 602 03 302 0

30 160 0 0 10

3

4

4 5

1

2

xij 1 2 3

26


06 1603 05 02 07 40

03 07 04 04 801 0

05 502 03 801 06 0

03 05 602 03 302 0

30 0 0 0 10

1

2

3

4

1 2 3 4 5xij

27


06 1603 05 02 07 40

03 07 04 04 801 0

05 502 03 801 06 0

03 05 602 03 302 0

30 0 0 0 10

3

4

4 5

1

2

xij 1 2 3

28


06 1603 05 02 07 40

03 07 04 04 801 0

05 502 03 801 06 0

03 05 602 03 302 0

30 0 0 0 10

1

2

3

4

1 2 3 4 5xij

29


306 1603 05 02 107 0

03 07 04 04 801 0

05 502 03 801 06 0

03 05 602 03 302 0

0 0 0 0 0

3

4

4 5

1

2

xij 1 2 3

30

• Differency parameter: difference between costs(𝑐𝑖𝑗) of components having the smallest and the-second-smallest unit cost remained in that rowor column

• Define the row or column which can be characterized by the highest differencyparameter and then from the selectedrow/column choose the elements, which havethe smallest unit costs

Vogel Initial Programming Method

200 ∗ 3 + 80 ∗ 1 + 10 ∗ 2 + 40 ∗ 3 + 80 ∗ 1 + 30 ∗ 3 + 20 ∗ 2 + 40 ∗ 2 = 1110

31


06 03 05 02 07 200 1

03 07 04 04 01 80 2

05 02 03 01 06 130 1

03 05 02 03 02 90 0

30 210 60 80 120

0 1 1 1 1

3

4

bj

ai

1

2

xij 1 2 3 4 5

32


06 03 05 02 07 200 1

-3 -7 -4 -4 801 0 -

05 02 03 01 06 130 1

03 05 02 03 02 90 0

30 210 60 80 40

2 1 1 1 4

2

3

xij 1 2 3 4

4

bj

5 ai

1

33


06 03 05 02 -7 200 1

-3 -7 -4 -4 801 0 -

05 02 03 01 -6 130 1

03 05 02 03 402 50 1

30 210 60 80 0

2 1 1 1 -

1

2

3

4

xij 1 2 3 4 5

34


-6 03 05 02 -7 200 1

-3 -7 -4 -4 801 0 -

-5 02 03 01 -6 130 1

303 05 02 03 402 20 1

0 210 60 80 0

- 1 1 1 -

1

2

3

4

bj

2 3 4 5 aixij 1

35


-6 03 05 -2 -7 200 2

-3 -7 -4 -4 801 0 -

-5 02 03 801 -6 50 1

303 05 02 -3 402 20 3

0 210 60 0 0

- 1 1 - -

4

1

2

3

xij 1 2 3 4 5

36


-6 03 05 -2 -7 200 2

-3 -7 -4 -4 801 0 -

-5 02 03 801 -6 50 1

303 -5 202 -3 402 0 -

0 210 40 0 0

- 1 2 - -

3 4 5xij 1 2

1

2

3

4

37


-6 2003 -5 -2 -7 0 -

-3 -7 -4 -4 801 0 -

-5 02 03 801 -6 50

303 -5 202 -3 402 0 -

0 10 40 0 0

- - -

4 5

1

xij 1 2 3

2

3

4

38


-6 2003 -5 -2 -7 0 -

-3 -7 -4 -4 801 0 -

-5 102 403 801 -6 0 -

303 -5 202 -3 402 0 -

0 0 0 0 0

- - - - -

4

bj

ai

1

2

3

xij 1 2 3 4 5

39


Distribution Method


Introduction

40

Potential System

𝑝𝑖𝑗 = 𝑐𝑖𝑗 − 𝑣𝑗 + 𝑢𝑖• 𝑝𝑖𝑗 parameters for the

tied elements, arealways 0

• Have to choose one rowor column arbitrary

• The order:– Red– Orange– Green

306 1603 05 02 107 0

03 07 04 04 801 6

05 502 03 801 06 1

03 05 602 03 302 5

6 3 7 2 7

4 5

2

xij 1 2 3

3

4

vj

ui

1

• In our example the solution of the Dantzig method is used

41

Potential System

𝑝𝑖𝑗 = 𝑐𝑖𝑗 − 𝑣𝑗 + 𝑢𝑖• The numbers: amount

of expected change of the sum (𝑍) value

• Choose the lowestelement which is below0

• If the lowest element is 0, or positive, then thesum will not decrease

-25 02 0

33 107 34 84 6

05 -33 06 1

23 75 63 5

6 3 7 2 7

1 2 3

4

vj

5 ui

1

2

3

4

42

Polygon on the matrix

• A polygon is need to setup on the matrix

• A corner is on thechosen element (now33)

• Other corners on tied elements

• Sign the corners by ‚+’ and ‚-’ alternately

• Start with ‚+’ and thechosen element

+ -

+

-

- +

1

2

3

4

30

80

xij 1 2 3 4 5

160

50

60 30

80

10

43

Current optimum

• The minimum of the ‚-’ signed elements is need tobe chosen

• Then this amount is needto add to the ‚+’ signedelements and extract fromthe ‚-’ signed ones

• The decrease of the 𝑍 is themultiplication of thechosen potential number, and the chosen amount

30 170

80

40 10 80

50 40

5

1

2

3

4

xij 1 2 3 4

30 ∗ 6 + 170 ∗ 3 + 80 ∗ 1 + 40 ∗ 2 + 20 ∗ 3 + 80 ∗ 1 + 50 ∗ 2 + 40 ∗ 2 = 1140

44

Potential System

306 1703 05 02 07 0

03 07 04 04 801 3

05 402 103 801 06 1

03 05 502 03 402 2

6 3 4 2 4

3

4

vj

1

2

4 5 uixij 1 2 3• The order:

– Red

– Orange

– Green

– Purple

– Blue

– Brown

45

Potential System

15 02 37 0

03 77 34 54 3

05 36 1

-13 45 33 2

6 3 4 2 4

3

4

vj

ui

1

2

1 2 3 4 5pij

46

Polygon on the matrix

- +

+

-

+ -4

5

1

2

3

xij 1 2 3 4

30 170

80

40 10 80

50 40

47

Current optimum

200

80

10 40 80

30 20 40

4 5

1

2

3

xij 1 2 3

4

48

Potential System

• The order:

– Red

– Orange

– Green

– Purple

– Blue

– Brown

06 2003 05 02 07 0

03 07 04 04 801 3

05 102 403 801 06 1

303 05 202 03 402 2

5 3 4 2 4

4

vj

5 ui

1

2

3

xij 1 2 3 4

49

Potential System

16 15 02 37 0

13 77 34 54 3

15 36 1

45 33 2

5 3 4 2 4

3 4 5 ui

1

pij 1 2

2

3

4

vj

50

Optimal solution

200

80

10 40 80

30 20 40

1 2 3 4 5

1

xij

2

3

4

51

Content

Excersises


Introduction

52

• Dummy supply or destination has needed to introduce

– Analyse the column and row summarises

– Introduce the supply OR destination what is needed

– The dummy supply or destination is an artificiallocation, which means that the needs are not fulfilled

• When a ij route is not feasible

– In the matrix usually signed by M

– In Solver a special condition is needed

Excersises – Practical Methods

53

• Conditions

– Row summarisies need to be equal

– Column summarisies need to be equal

– Nonnegativity cretiria

– Special criteria for infeasible routes

• Nonnegativity: 𝑥𝑖𝑗 ≥ 0 ∀𝑖, 𝑗

• Special criteria: 𝑥𝑖𝑗 ≤ 0 for infeasible routes

• If 𝑥𝑖𝑗 ≥ 0 and 𝑥𝑖𝑗 ≤ 0, then 𝑥𝑖𝑗 = 0 for infeasible routes

Excersises – Solver

BME FACULTY OF TRANSPORTATION ENGINEERING AND VEHICLE ENGINEERING32708-2/2017/INTFIN COURSE MATERIAL SUPPORTED BY EMMI

Dr. Tibor SIPOS Ph.D.Dr. Árpád TÖRÖK Ph.D.Zsombor SZABÓ

email: [email protected]

BUDAPEST UNIVERSITY OF TECHNOLOGY AND ECONOMICS

transportation problemkukg.bme.hu/.../2019/04/07_transportation-problem.pdf2019/04/07 · northwest...

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